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agasfer [191]
3 years ago
6

Xyxyydfufggivivihogcufuf​

Engineering
1 answer:
Genrish500 [490]3 years ago
7 0

Answer:

ummm why is you doing this

Explanation:

It doesnt make sense.

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A motorcycle starts from rest with an initial acceleration of 3 m/s^2, and the acceleration then changes with the distance s as
katrin2010 [14]

Answer:

Follows are the solution to this question:

Explanation:

Calculating the area under the curve:  

A = as

   =\frac{1}{2}(3 +6 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(6+4 \frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(9 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(10\frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(900 \frac{m^2}{s^2})+ \frac{1}{2}(1,000\frac{m^2}{s^2}) \\\\=(450 \frac{m^2}{s^2})+ (500\frac{m^2}{s^2}) \\\\= 950 \ \frac{m^2}{s^2}

Calculating the kinematics equation:

\to v^2 = v^2_{o} + 2as\\\\

        =0+ \sqrt{2as}\\\\ = \sqrt{2(A)}\\\\= \sqrt{2(950 \frac{m^2}{s^2})}\\\\= 43.59 \frac{m}{s}

Calculating the value of acceleration:  

\to a= \frac{dv}{dt}

=\frac{dv}{ds}(\frac{ds}{dt}) \\\\=v\frac{dv}{ds}\\\\\to \frac{dv}{ds}=\frac{a}{v}

\to \frac{dv}{ds} =\frac{4 \frac{m}{s^2}}{43.59 \frac{m}{s}} \\\\

         =\frac{0.092}{s}

3 0
2 years ago
Two uniformly charged conducting plates are parallel to each other. They each have area A. Plate #1 has a positive charge Q whil
Karo-lina-s [1.5K]

Answer:

E = \frac{3Q}{2A\epsilon_0}

Explanation:

By Gauss Law for electric field:

E = \frac{\sigma}{2\epsilon_0}

Where \sigma is the charge density Q/A. Since we have 2 parallel  plates with different charge, the electric field at point P in the gap would be the sum of 2 field

E = E_1 + E_2

E = \frac{Q}{2A\epsilon_0} + \frac{2Q}{2A\epsilon_0}

E = \frac{3Q}{2A\epsilon_0}

5 0
3 years ago
Problem 9.11 A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain f
BabaBlast [244]

Answer:

the critical flaw length is 10.06 mm

Explanation:

Given the data in the question;

plane strain fracture toughness K_{tc = 92 Mpa√m

yield strength σ_y = 900 Mpa

design stress is one-half of the yield strength ( 900 Mpa / 2 ) 450 Mpa

Y = 1.15

we know that;

Critical crack length a_c = 1/π( K_{tc / Yσ )²

we substitute

a_c = 1/π( 92 Mpa√m / (1.15 × 450 Mpa  )²

a_c = 1/π( 92 Mpa√m / (517.5 Mpa  )²

a_c = 1/π( 0.177777  )²

a_c = 1/π( 0.03160466 )

a_c = 0.01006 m = 10.06 mm

Therefore, the critical flaw length is 10.06 mm

{ a_c = ( 10.06 mm ) > 3 mm

The critical flow is subject to detection

5 0
3 years ago
What is Class Night in Puerto Rico?
valentina_108 [34]

Answer:

b

Explanation:

8 0
3 years ago
The sticker inside the door of my car says that the tire pressure should be 32 psig (322 kPa) when the tire is cold. Before a ro
Neko [114]

Answer:

37 psi

Explanation:

For ideal gases this equation applies:

p1*V1/T1 = p2*V2/T2

Since we are assuming volume remains constant:

V2 = V1

p1/T1 = p2/T2

p2 = p1*T2/T1

The temperatures must be in absolute scale.

T1 = 15 + 273  = 288 K

T2 = 60 + 273 = 333 K

Then:

p2 = 32 * 333 / 288 = 37 psi

7 0
3 years ago
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